Wheels and Friction Part 2

As promised, I will discuss how wheels under friction arrive at the steady state I talked about in part 1. I must thank Bihui for alerting me to this website, which allows tex equation rendering and enabled me to write this unfortunately technical entry. Those of you less interested in gory mathematical detail should stay tuned for the next issue. For now, we start with a stationary wheel on a flat surface. To move this wheel, we must apply a force. Once the wheel is moving, we have constant kinetic friction if there is relative motion between the bottom of the wheel and the surface, and static friction equal and opposite to the applied force up to some maximum if there is no such relative motion. In other words,

\begin{eqnarray} 
f = \left\{ \begin{array}{cc}
-f_{kinetic} & R \theta'[t] < x'[t] \\
f_{kinetic} & R \theta'[t] > x'[t] \\
-Sgn[Fa]Min[|Fa|,f'_{max}] & R \theta'[t] = x'[t]
\end{array} \right\}
\end{eqnarray}

where f is friction, Fa is the applied force, R is the wheel's radius, ɵ is the wheel's angle, x is the wheel's horizontal position, and t is time. ɵ and x have equations of motion given by Newton's 2nd law:

m x''[t] = f[t] + Fa[t]
I \theta''[t] = - R f[t] + ra[t] Fa[t]

where m is the wheel's mass, I = 1/2 m R^2 is the wheel's moment of inertia, and ra is the normal signed distance between the applied force and the wheel's center. We can non-dimensionalize these equations by defining

X=x/R
r=ra/R
F=Fa/f_{kinetic}
f_{max} = f'_{max}/f_{kinetic} 
T=t \sqrt{\frac{f_{kinetic}R}{m}}

to obtain:

X''[T] = F[T] + H[\theta'[T] - X'[T]]
\theta''[T] = 2 (r[T] F[T] - H[\theta'[T] - X'[T]])
\begin{eqnarray} 
H[y-z] = \left\{ \begin{array}{cc}
-1 & y < z \\
1 & y > z \\
-Sgn[F[T]]Min[|F[T]|,f_{max}] & y = z
\end{array} \right\}
\end{eqnarray}

Note that when ɵ'[T] = X'[T], |F[T]| need only be non-zero but not greater than the maximum static friction to affect the wheel, since ɵ''[T] need not be 0. Unlike non-circular objects, a wheel can move even when static friction exceeds the applied force!

Consequently, we can actually disregard static friction, because of a confusing mathematical technicality you should skip over unless you doubt me. For non-zero F[T], ɵ'[T] = X'[T] at only isolated T, which can be disregarded in solving the equations of motion. When F[T]=0 and ɵ'[T]=X'[T], friction is static and 0. So without affecting the solution, we can write

\begin{eqnarray} 
H[y-z] = \left\{ \begin{array}{cc}
-1 & y < z \\
1 & y > z \\
0 & y = z
\end{array} \right\}
\end{eqnarray}

The solution to the equations of motion can be obtained by time integration over pieces where ɵ'[T] <, > or = X'[T]. If F[T]=0 for T greater than some value, X''[T] = H[ɵ'[T] - X'[T]] and ɵ''[T] = -2 H[ɵ'[T] - X'[T]]. So X'[T] decreases while ɵ'[T] increases if X'[T] > ɵ'[T] and vice versa if X'[T] < ɵ'[T], until ɵ'[T] = X'[T] and X''[T]=ɵ''[T]=0. Therefore, the wheel always shifts to a steady state when there is no applied force.

With any steady state a natural question to ask is one about its stability. This state could be said to have neutral stability, in the sense that a perturbation in X'[T] or ɵ'[T] (via a non-zero F) causes the wheel to reach a different steady state with new equal X'[T] and ɵ'[T].

I can't squeeze anymore tidbits from this mathematical analysis (maybe you can), which is admittedly rich in technicality and correspondingly low on understanding. For a better picture of what's going on, you can try the Mathematica notebook advertised in the previous entry, where the equations of motion are solved by finite difference. Or just stay tuned for part 3, where I'll consider some simple cases. Phew!